$\Sigma M_D = 0$
$5R_1 = 50(0.5) + 25$
$R_1 = 10 \, \text{kN}$
$\Sigma M_A = 0$
$5R_2 + 25 = 50(4.5)$
$R_2 = 40 \, \text{kN}$
To draw the Shear Diagram
- V_{A} = R_{1} = 10 kN
- V_{B} = V_{A} + Area in load diagram
V_{B} = 10 + 0 = 10 kN
- V_{C} = V_{B} + Area in load diagram
V_{C} = 10 + 0 = 10 kN
- V_{D} = V_{C} + Area in load diagram
V_{D} = 10 - 10(3) = -20 kN
V_{D2} = -20 + R_{2} = 20 kN
- V_{E} = V_{D2} + Area in load diagram
V_{E} = 20 - 10(2) = 0
- Solving for x:
x / 10 = (3 - x) / 20
20x = 30 - 10x
x = 1 m
To draw the Moment Diagram
- M_{A} = 0
- M_{B} = M_{A} + Area in shear diagram
M_{B} = 0 + 1(10) = 10 kN·m
M_{B2} = 10 - 25 = -15 kN·m
- M_{C} = M_{B2} + Area in shear diagram
M_{C} = -15 + 1(10) = -5 kN·m
- M_{x} = M_{C} + Area in shear diagram
M_{x} = -5 + ½(1)(10) = 0
- M_{D} = M_{x} + Area in shear diagram
M_{D} = 0 - ½(2)(20) = -20 kN·m
- M_{E} = M_{D} + Area in shear diagram
M_{E} = -20 + ½ (2)(20) = 0